Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Proving Trigonometric Identities


Date: 02/12/99 at 13:47:56
From: Hannah Calsyn
Subject: Trig Identities

Can you teach me how to prove trig identities? I don't know where to 
start or what to do exactly.


Date: 02/12/99 at 15:59:29
From: Doctor Rob
Subject: Re: Trig Identities

Thanks for writing to Ask Dr. Math!

There are usually many ways to prove trigonometric identities. Some 
are short and very elegant. Others are longer and more tedious, but 
any proof should do.

The shortest proofs involve *pattern recognition*. In the equation you
are trying to prove, you detect a pattern that appears in one of the
standard identities you know, and that allows you to make a 
substitution which simplifies the identity a lot. There is a knack to
this which is a bit hard to teach, but it comes with practice. See the
following web page for a list of standard identities:   
   http://mathforum.org/dr.math/faq/formulas/faq.trig.html   

There is a systematic way to produce a proof, which may be fairly 
long, but still valid. I will describe it below.

A proof of an identity is often constructed by starting with one
expression on the lefthand side of the equation and another on the
righthand side which you want to prove as equal. You make 
substitutions on both sides until they are both reduced to the same 
identical expression. Then, since the last equation is identically 
true, you can reverse the steps to conclude that the first equation 
that you were trying to prove is also true.

Alternatively, you can rearrange your work to start with the lefthand
side of the first equation, work down the chain of equal expressions
on the lefthand sides until you reach the bottom, then work up the
chain of equal expressions on the righthand side until you are back at
the top.

Here is an example:

   cos^4(x) - sin^4(x) = cos(2*x)
   [cos^2(x) - sin^2(x)]*[cos^2(x) + sin^2(x)] = 2*cos^2(x) - 1

Here we factored the lefthand side as the difference of two squares, 
and we used the cosine double-angle formula on the righthand side.

   [cos^2(x) - sin^2(x)]*1 = 2*cos^2(x) - 1

Here we used the identity cos^2(x) + sin^2(x) = 1.

   cos^2(x) - [1 - cos^2(x)] = 2*cos^2(x) - 1

Here we used 1 - cos^2(x) = sin^2(x).

   2*cos^2(x) - 1 = 2*cos^2(x) - 1

Here we just expanded and combined like terms. Now we have an 
obviously true equation at the bottom. To prove the original identity, 
we can just reverse the steps,

   (1.)  2*cos^2(x) - 1 = 2*cos^2(x) - 1
   (2.)  cos^2(x) - [1 - cos^2(x)] = 2*cos^2(x) - 1
   (3.)  cos^2(x) - sin^2(x) = 2*cos^2(x) - 1
   (4.)  [cos^2(x) - sin^2(x)]*[cos^2(x) + sin^2(x)] = 2*cos^2(x) - 1
   (5.)  cos^4(x) - sin^4(x) = cos(2*x)

and supply the reasons. Alternatively, we can use this form:

   cos^4(x) - sin^4(x) = [cos^2(x)] - sin^2(x)]*[cos^2(x) + sin^2(x)]
                       = cos^2(x) - sin^2(x)
                       = cos^2(x) - [1 - cos^2(x)]
                       = 2*cos^2(x) - 1
                       = cos(2*x)

by working down the left and up the right of our original derivation, 
and supply the reasons.

Here is a systematic way to produce a trigonometry proof:

1. Convert all cosecants, secants, cotangents, and tangents to 
expressions involving only sines and cosines. The first group of five 
identities on the above web page will allow you to do this easily.  
That will give you a "simpler" equation to prove.

2. Look at all the angles in the sines and cosines in the new equation
you are trying to prove. If any is a sum or difference of angles, use
a sine or cosine sum-of-angles formula, as in the 9th and 10th groups
of identities from the Dr. Math FAQ web page. That will give you a
"simpler" equation to prove.

3. Look at all the angles in the sines and cosines in the new equation
you are trying to prove. If any is a multiple of some angle, use a 
sine or cosine multiple-angle formula, as in the 11th through 16th 
groups of identities from the web page. Likewise for half-angles and 
the 12th group of identities. That will give you a "simpler" equation 
to prove.

4. Expand all the expressions in sight, combine like terms, and 
simplify. That will give you a "simpler" equation to prove.

5. Replace all occurrences of the square or higher power of a cosine
using the identity cos^2(u) = 1 - sin^2(u). Expand, combine, and
simplify again. That will give you a "simpler" equation to prove.

6. Factor numerator and denominator if possible and cancel common 
factors if any. That will give you a "simpler" equation to prove.

7. At this point, you should have an equation that is obviously true, 
of the form E = E, where E is some expression involving sines to 
powers and possibly some cosines to the first power.

8. From this sequence of equations, construct a proof of the original
equation using one of the two methods described above: either working
from the bottom up, or else working from the top left down, over, and 
up, to the top right.

I did warn you that this was tedious! It will work, however, in almost
all cases, and with minor variations in essentially all cases. Be sure
to do the algebra correctly when you are doing "expand, combine,
simplify, and factor" operations!

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/